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Aqa chemistry unit 1 unofficial mark scheme

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I believe there was also 2 marks for stating the definitions of saturated and hydrocarbon in terms of something?


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Original post by ALevelBro
That's a different molecule:

Mg3N2 is Magnesium (element) bonding with Nitrogen (also an element)...

You were asked what happens when Magnesium (element) bonds with the N3- (compound).


crap. thanks
just so pressed for time these silly mistakes happen
Reply 82
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Original post by Zer0.
Hi. I'm just inquiring regarding all of the calculation questions. I converted the masses to KG (As per the SI Units); I'm just wondering whether this will affect my marks.

Thanks in advance @PS4


I think your answers would be 1000 times less than the answers that will be on the mark scheme. This is because moles = mass (g) / Mr (g/mol). This means that you used mass as kg but mr as g/mol so yout answers would be 1000 times less than mine. However, you would only lose one mark as e.c.f marks would be awarded.
Original post by Louisss
I believe there was also 2 marks for stating the definitions of saturated and hydrocarbon in terms of something?


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Yeah, I wonder what those were? Saturated - double bond, hydrocarbon - chain of hydrogen and carbons?
Original post by Louisss
I believe there was also 2 marks for stating the definitions of saturated and hydrocarbon in terms of something?


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This.

Saturated - Contains a only C-C single bonds (No double C=C bonds)
Hydrocarbons - Contains carbon and hydrogen atoms only
Reply 85
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Original post by Louisss
I believe there was also 2 marks for stating the definitions of saturated and hydrocarbon in terms of something?


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THANK YOU SOOOO much hahaha. I have been going mad trying to think of the last few marks :biggrin:
Thought what?
53 for an a?
Original post by zakiahmed
If i rpunded to 3sf for the 190g question will i lose a mark? Also woupd i gain one for saying argon has a full outee shell so its more difficult to remove an electeon from there? And for the solid crystal question i talked about filtration too.


I believe the gas question didnt specify how many sig figs, so this should be correct (probably a range of accepted answers). For argon i put largest nuclear charge, similar shielding. I dont think full shell of electrons will suffice.
For the water question, this was water of crystallisation - trapped in the molecule. Therefore, it cant be filtered, only evaporated. I think the correct answer was heat until mass is constant.
Original post by Connorbwfc
You will either lose 1 or 2. By the way you were meant to divide by 0.95, not times by 1.05.


For this question:
It asks to work out 550g which is 95% yield. You don't divide 550g to get 95%, you leave it like that because that's the amount you get assuming a 95% yield.
It's 550/65 which is 8.461 moles multiply 2 to get moles of NaNH2 which is 16.922 moles:
Multiply by 39 (Mr of NaNH2) to get 659.958.

Then you calculate 95% of the 659.958 because this is theoretical yield but you want actual yield so you multiply by 0.95 to get something like 626g to three sig fig.
Look at a similar question (3di in the JAN 2011 paper)
JAN 2011.png223.8KB
QUESTION 3Dii.png191.1KB
zzzz.png

remember guys, what you get in unit 2 really matters
nearly accounts to 50% of your overall grade so if you flop that...yeh

look at the pic to see what I mean
Reply 91
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Original post by zakiahmed
I just said plastics for the polymer question. You yhink thats ok


I am not sure sorry. I wrote poly(propene) which is formed by the additional polymerisation of propene and plastics could also be made from propene.
I think it should be accepted but it did ask for specifically propene so it should be poly(propene)
Original post by chemistrypower
For this question:
It asks to work out 550g which is 95% yield. You don't divide 550g to get 95%, you leave it like that because that's the amount you get assuming a 95% yield.
It's 550/65 which is 8.461 moles multiply 2 to get moles of NaNH2 which is 16.922 moles:
Multiply by 39 (Mr of NaNH2) to get 659.958.

Then you calculate 95% of the 659.958 because this is theoretical yield but you want actual yield so you multiply by 0.95 to get something like 626g to three sig fig.
Look at a similar question (3di in the JAN 2011 paper)


I hope what youre saying is right
this is exactly what I did
Original post by chemistrypower
For this question:
It asks to work out 550g which is 95% yield. You don't divide 550g to get 95%, you leave it like that because that's the amount you get assuming a 95% yield.
It's 550/65 which is 8.461 moles multiply 2 to get moles of NaNH2 which is 16.922 moles:
Multiply by 39 (Mr of NaNH2) to get 659.958.

Then you calculate 95% of the 659.958 because this is theoretical yield but you want actual yield so you multiply by 0.95 to get something like 626g to three sig fig.
Look at a similar question (3di in the JAN 2011 paper)


This is exactly what i did. However the wording of the question threw me, as i wasnt sure whether the 550g WAS the 95% yield or whether we had to work out what mass of reactants to produce 95% of 550g.
Original post by chemistrypower
For this question:
It asks to work out 550g which is 95% yield. You don't divide 550g to get 95%, you leave it like that because that's the amount you get assuming a 95% yield.
It's 550/65 which is 8.461 moles multiply 2 to get moles of NaNH2 which is 16.922 moles:
Multiply by 39 (Mr of NaNH2) to get 659.958.

Then you calculate 95% of the 659.958 because this is theoretical yield but you want actual yield so you multiply by 0.95 to get something like 626g to three sig fig.
Look at a similar question (3di in the JAN 2011 paper)


This is a different type of question. In the question we got given you were asked to find amount of the reactants given that yield so you had to work back. In this question you are working out things about the products only.
Original post by mickel_w
This is exactly what i did. However the wording of the question threw me, as i wasnt sure whether the 550g WAS the 95% yield or whether we had to work out what mass of reactants to produce 95% of 550g.



550 Grams was the 95% yield I think.
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Original post by chemistrypower
For this question:
It asks to work out 550g which is 95% yield. You don't divide 550g to get 95%, you leave it like that because that's the amount you get assuming a 95% yield.
It's 550/65 which is 8.461 moles multiply 2 to get moles of NaNH2 which is 16.922 moles:
Multiply by 39 (Mr of NaNH2) to get 659.958.

Then you calculate 95% of the 659.958 because this is theoretical yield but you want actual yield so you multiply by 0.95 to get something like 626g to three sig fig.
Look at a similar question (3di in the JAN 2011 paper)




We are working backwards in this question though. In the question in Jan 2011, you find the maximum mass and then times by 0.95 to get the actual mass. However, we were giving the actual mass produced.

% yield = (Actual yield/Maximum yield) times 100

95 = (550 / Maximum yield) times 100

0.95 = 550/ Maximum yield.

Therefore the maximum yield is 550/0.95 which equals 578.9g
Original post by Connorbwfc
This is a different type of question. In the question we got given you were asked to find amount of the reactants given that yield so you had to work back. In this question you are working out things about the products only.

Damn. Its annoying, because if i knew exactly what it was asking i could have worked it out correctly.
How many marks do you reckon i could get for that question? E.c.f.?
Original post by mickel_w
Damn. Its annoying, because if i knew exactly what it was asking i could have worked it out correctly.
How many marks do you reckon i could get for that question? E.c.f.?


Probably lose 1. Maybe 2 if you're unlucky.
around what mark would you need to get full UMS in Unit 2?

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